For making $2$ mixed teams you'll need $2$ females and $2$ males.
You have $4\choose 2$ ways to choose the $2$ males and $3\choose 2$ ways to choose the $2$ females.
Now you can pair them in two different ways since each female can go with either male.
So that's $2\times {4\choose 2} \times {3\choose 2}$
EDIT for better understanding:
Note that $2\times {4\choose 2} \times {3\choose 2}=36$
Now, let's solve the problem somehow different.
There are teams A and B on each tennis game.
For team A we've got $4$ males and $3$ females, so $4\times 3=12$ possible A teams.
For team B we've got $3$ remaining males and $2$ remaining females, so $3\times 2=6$ possible B teams.
So we have $12\times 6=72$ ways to create a A & B team. But taking into account that team A & B are indistinguishable, we have to divide it by $2$ (Albert & Berta vs Charles & Diana is the same that Charles & Diana vs Albert & Berta). And $\frac{72}{2}=36$. Calculated differently, but as we see is the very same number.
As pointed out by Mr. Shore and Mr. Taussig, The instructions of the questions allows us to perform the calculation of "Including $\text{B}$ but not $\text{A}$", so We have $16 + 6 = 22$ ways of choosing.
Thanks for clearing my doubt on this one.
Best Answer
Methods 1 and 2 should work.
Method 3 does not correctly count the set. You would be counting M1 + M2W1W2 as a different arrangement than M2 + M1W1W2 even though they represent the same team.